What is the integral of sqrt(x^4+x^2) ?

The integral of sqrt(x^4+x^2) is 1/3 (x^2+1)^{3/2}+C

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integral of sqrt(x^4+x^2)

\int x^{4} + x^{2} d x

\frac{1}{3} (x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{4} + x^{2} d x

Simplify

x^{4} + x^{2} : x^{2} x^{2} + 1

= \int x^{2} x^{2} + 1 d x

x^{2} = (x),

assuming

x \geq 0

= \int x x^{2} + 1 d x

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = x^{2} + 1

= \frac{( x ^{2} + 1 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 1 ) ^{3}}{3} : \frac{1}{3} (x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{3} (x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (x^{2} + 1)^{\frac{3}{2}} + C

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What is the integral of sqrt(x^4+x^2) ?
The integral of sqrt(x^4+x^2) is 1/3 (x^2+1)^{3/2}+C